Question: Simplify and expand the following expression: $ \dfrac{a}{a + 1}+\dfrac{3a + 1}{a + 4} $
Explanation: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(a + 1)(a + 4)$ Multiply the first term by $\dfrac{a + 4}{a + 4}$ $ \begin{align*} \dfrac{a}{a + 1} \times \dfrac{a + 4}{a + 4} & = \dfrac{(a)(a + 4)}{(a + 1)(a + 4)} \\ & = \dfrac{a^2 + 4a}{(a + 1)(a + 4)}\end{align*} $ Multiply the second term by $\dfrac{a + 1}{a + 1}$ $ \begin{align*} \dfrac{3a + 1}{a + 4} \times \dfrac{a + 1}{a + 1} & = \dfrac{(3a + 1)(a + 1)}{(a + 4)(a + 1)} \\ & = \dfrac{3a^2 + 4a + 1}{(a + 4)(a + 1)}\end{align*} $ Now we have: $ = \dfrac{a^2 + 4a}{(a + 1)(a + 4)} + \dfrac{3a^2 + 4a + 1}{(a + 4)(a + 1)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{a^2 + 4a + 3a^2 + 4a + 1}{(a + 1)(a + 4)} $ $ = \dfrac{4a^2 + 8a + 1}{(a + 1)(a + 4)}$ Expand the denominator: $ = \dfrac{4a^2 + 8a + 1}{a^2 + 5a + 4}$